Question: Simplify the following expression and state the condition under which the simplification is valid: $a = \dfrac{n^2 - 3n}{n^2 - 9}$
Solution: First factor the expressions in the numerator and denominator. $ \dfrac{n^2 - 3n}{n^2 - 9} = \dfrac{(n)(n - 3)}{(n + 3)(n - 3)} $ Notice that the term $(n - 3)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(n - 3)$ gives: $a = \dfrac{n}{n + 3}$ Since we divided by $(n - 3)$, $n \neq 3$. $a = \dfrac{n}{n + 3}; \space n \neq 3$